In Maxima, how can I solve this equation?

Question

In Maxima, how can I solve this equation? And actually I have 24 unknown value with lots of this kind of equation. How I can use Maxima to solve?

to know the unknown value: a, b, c, d

eq:[a*c=8,a*d=10,b*c=12,b*d=15,a+b=5,c+d=9];
solve(eq,[a,b,c,d]);

Do you have more equations than unknowns? Are there any constraints on the variables? e.g. that the variables must be positive or must be integers or some other constraint. — Robert Dodier, Sep 01 '17 at 16:51
I have more equations than unknown for sure, but there is still another problem is that I might have more than 1 equation equal to the same value.For the constraint of valuables, I think it must be possitive. Take another example: unknow values are belong to 3 groups (a,b.c),(d,e,f),(g,h,i) , each equation is construct by choose one unknown value in each group. a+b+c=1 d+e+f=1 g+h+i=1 a*d*g=0.3 b*e*h=0.2 c*f*i=0.6 b*d*g=0.36 — Yu Hsien Lin, Sep 01 '17 at 19:57
By the way, you can enclose code in backticks so that it is formatted specially. E.g. `a*d*g` in backticks versus a*d*g without backticks. — Robert Dodier, Sep 02 '17 at 04:55

score 1 · Answer 1 · answered Sep 03 '17 at 08:28

With four unknowns you have to have four equations. Ignore one set of equations and run:

sol1: solve([a*c=8,a*d=10,b*c=12,a+b=5],[a,b,c,d]);
              [[a = 2, b = 3, c = 4, d = 5]]

Then, ignore an other set of equations and run again:

sol2: solve([a*c=8,a*d=10,b*c=12,c+d=9],[a,b,c,d]);
              [[a = 2, b = 3, c = 4, d = 5]]

If the two results are the same, then your system is consistent and has solution sol1 ≡ sol2.

score 0 · Answer 2 · answered Sep 02 '17 at 05:26

Your equations are, if I'm not mistaken, polynomials in several variables of degree 2 or greater (where the degree is equal to the maximum number of variables multiplied together). You say that you have more equations than unknowns.

If I'm not mistaken, in general there are no exact solutions, but you can look for a best-fitting solution via least squares. I don't know how many locally best-fitting solutions there might be. You might ask on math.stackexchange.com about the number of solutions and other characteristics of such equations.

Here is a way to look for a locally best-fitting solution in Maxima. Form a "figure of merit" expression (fom below) from the equations, and look for a local minimum of it. I'll use lbfgs to search for a numerical approximation.

(%i3) eq:[a*c=8,a*d=10,b*c=12,b*d=15,a+b=5,c+d=9];
(%o3) [a*c = 8,a*d = 10,b*c = 12,b*d = 15,b+a = 5,d+c = 9]
(%i4) eq0 : map (lambda ([e], lhs(e) - rhs(e)), eq);
(%o4) [a*c-8,a*d-10,b*c-12,b*d-15,b+a-5,d+c-9]
(%i5) eq2 : expand (eq0^2);
(%o5) [a^2*c^2-16*a*c+64,a^2*d^2-20*a*d+100,b^2*c^2-24*b*c+144,
       b^2*d^2-30*b*d+225,b^2+2*a*b-10*b+a^2-10*a+25,
       d^2+2*c*d-18*d+c^2-18*c+81]
(%i6) load (lbfgs);
(%o6) "/usr/local/share/maxima/branch_5_40_base_95_g4c9cbbd/share/lbfgs/lbfgs.mac"

(%i9) fom : apply ("+", eq2);
(%o9) b^2*d^2+a^2*d^2+d^2+2*c*d-30*b*d-20*a*d-18*d+b^2*c^2
             +a^2*c^2+c^2-24*b*c-16*a*c-18*c+b^2+2*a*b-10*b+a^2
             -10*a+639

Here fom is just the sum of squares of error (i.e. the discrepancy between the right hand side and left hand sides of the equations, squared and added up).

(%i10) lbfgs (fom, [a, b, c, d], [1, 1, 1, 1]/4, 1e-4, [1, 0]);
*************************************************
  N=    4   NUMBER OF CORRECTIONS=25
       INITIAL VALUES
 F=  6.198906250000000D+02   GNORM=  4.916696680699350D+01
*************************************************

   I  NFN     FUNC                    GNORM                   STEPLENGTH

   1    5     1.029984221465989D+01   3.303822830240210D+01   1.342303790326981D-01  
   2    7     1.858590097968602D+00   7.066597977731433D+00   2.567282877401947D-01  
   3    9     1.707935938873334D+00   7.250276888460332D+00   1.939008057328834D-01  
   4   10     1.071933775239074D+00   3.590373568045903D+00   1.000000000000000D+00  
   5   11     6.984083995048138D-01   2.736311274794537D+00   1.000000000000000D+00  
   6   12     2.756317886006059D-01   2.992718120186393D+00   1.000000000000000D+00  
   7   13     4.947211249555039D-02   2.105297770017910D+00   1.000000000000000D+00  
   8   14     8.047123928918154D-04   2.073845793942183D-01   1.000000000000000D+00  
   9   15     7.401148110375289D-06   2.195111965609954D-02   1.000000000000000D+00  
  10   16     3.129228502984915D-07   4.565302248467960D-03   1.000000000000000D+00  
  11   17     1.750777300912887D-09   4.710495250075657D-04   1.000000000000000D+00  

 THE MINIMIZATION TERMINATED WITHOUT DETECTING ERRORS.
 IFLAG = 0
(%o10) [a = 2.000003188592,b = 2.999996386563707,
        c = 4.000008973580663,d = 5.00000040529949]

I can guess that the a = 2, b = 3, c = 4, d = 5 might be an exact solution. I can verify that:

(%i11) subst ([a = 2, b = 3, c = 4, d = 5], fom);
(%o11) 0

That's encouraging in this case, but in general I think there won't be an exact solution near the approximate one. The approximate solution may be the best you can do.

Hi, Thanks a lot for your explanation. But, when I use this way to my real data, the result still give me negative value which is not correct , how can I add some limitations to the best fitting value,e.g limit values lower than 1 and a possitive value, because in my case ,each unknown values means one possibility. — Yu Hsien Lin, Sep 04 '17 at 13:07
Well, constraints make the problem quite a lot more complex. There is software such as COBYLA (a web search will find references) which handles constrained optimization. There is a Maxima interface for COBYLA; `?? cobyla` at the input prompt will show the documentation for it. There are many other non-Maxima packages for constrained optimization as well which you can look at. It might be possible to handle some simple constraints without resorting to special packages, e.g. by replacing a variable c with 1 - a - b to represent a + b + c = 1. — Robert Dodier, Sep 04 '17 at 16:20

In Maxima, how can I solve this equation?

2 Answers2